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I got stuck with the following question:

Consider the following heuristic: Start with a tour containing only one vertex. At each step, find the vertex outside the tour with the lesser distance to some vertex of the tour. Let v be the outter vertex and u be the inner vertex. Add v right after u in the tour. Suppose that your edges follow the triangle distance property. How can we show that this heuristic is a 2-approximation for the TSP-metric problem?

Does anyone know how to start that?

Thanks in advance

gcolucci
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  • I really can't see why someone would vote this down... – gcolucci Jun 16 '14 at 03:48
  • Hint: What mathematical proof technique fits a situation where you have a state that you can assume has some property, and you incrementally modify this state somehow, and want to show that the resulting state continues to have this property? – j_random_hacker Jun 16 '14 at 04:01
  • (I'm not certain that this technique will work here, but it's the obvious one to try first. And I don't think you should have been downvoted, but it's currently fashionable at SO to interpret the rules as narrowly as possible, and help with a proof is more more mathematics/CS than programming in some people's minds. You could also try your question at the CS SE website.) – j_random_hacker Jun 16 '14 at 04:05
  • Hi @j_random_hacker, I understand you're suggesting to prove it by induction. If I have the algorithm working for `n-1` vertex, when I add a new one, I think I don't have too much control of how the previous tour will rearrange itself to include the nth vertex... – gcolucci Jun 16 '14 at 04:07
  • Right, I really don't understand why there's people who just downvote the question and give no reason to it. I mean, there's a chance of other people lose the interest in seeing a downvoted question an the main goal of the website, which is to spread knowledge gets lost. But anyway, thanks for the advice, I posted it in the Mathematics as well. – gcolucci Jun 16 '14 at 04:10
  • There is some (unknown to us) *optimal* solution for n vertices, and in that solution, including vertex n adds some distance x (i.e. deleting vertex n would reduce the total tour distance by x to some smaller value y). We know that x >= 0 because otherwise the edge lengths would violate the Triangle Inequality. See if you can relate x and y to the penalty incurred by adding vertex n to the heuristic solution for n-1 vertices. – j_random_hacker Jun 16 '14 at 04:19
  • Why do you believe that the proposed algorithm yields a 2-approximation? Is this homework? – Codor Jun 16 '14 at 08:47
  • "I posted it [\[on\] Mathematics](http://math.stackexchange.com/questions/835602/2-approximation-for-tsp-metric) as well" - [Is cross-posting a question on multiple Stack Exchange sites permitted if the question is on-topic for each site?](http://meta.stackexchange.com/q/64068) – Bernhard Barker Jun 16 '14 at 09:43

1 Answers1

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Apparently, it is impossible to prove, the described algorithm is not a 2-approximation. The Wikipedia article mentions the publication

Rosenkrantz, Daniel J.; Stearns, Richard E.; Lewis, Philip M., II (1977), "An Analysis of Several Heuristics for the Traveling Salesman Problem", SIAM Journal on Computing 6 (5): 563–581, doi:10.1137/0206041

in which apparently the authors show that the Nearest Neighbor heuristic yields an approximation ratio of Theta( log n ), where n is the number of locations, even if the instance satisfies the triangle inequality:

Rosenkrantz et al. [1977] showed that the NN algorithm has the approximation factor Theta(log|V|) for instances satisfying the triangle inequality.

However, the OP might have described a different algorithm; an analysis of the approximation ratio of different greedy heuristics might be found in the followin article.

SIAM Journal on Computing, 1977, Vol. 6, No. 3 : pp. 563-581 An Analysis of Several Heuristics for the Traveling Salesman Problem Rosenkrantz, D., Stearns, R., and Lewis, II, P. (doi: 10.1137/0206041)

Codor
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  • Hi @Codor - thanks for the help. Indeed, this is homework, but I'm starting to think that it's misspecified. Also, I think it's a bit different from the NN algorithm, since the latter includes the closest vertex always at the end of the cycle, not between other two, as far as I know – gcolucci Jun 16 '14 at 15:57
  • @user137227 Thanks for the clarification so far. If you obtain a definite greedy algorithm with approximation ratio 2, please post an answer to this question; I would be keen to hear about it. – Codor Jun 16 '14 at 17:54
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    @user137227 Please see my updated answer; perhaps the algorithm is analyzed in the included reference - however I have no access to the article. – Codor Jun 16 '14 at 18:12